Ingenieria economica

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SOLUTIONS TO SELECTED PROBLEMS Student: You should work the problem completely before referring to the solution.

CHAPTER 1
Solutions included for problems 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, and 49 1.1 Time value of money means that there is a certain worth in having money and the worth changes as a function of time. 1.4 Nearest, tastiest, quickest, classiest, mostscenic, etc 1.7 Minimum attractive rate of return is the lowest rate of return (interest rate) that companies or individuals consider to be high enough to induce them to invest their money. 1.10 Rate of increase = [(29 – 22)/22]*100 = 31.8% 1.13 Profit = 8 million*0.28 = $2,240,000 1.16 (a) Equivalent future amount = 10,000 + 10,000(0.08) = 10,000(1 + 0.08) = $10,800 (b) Equivalent past amount: P+ 0.08P = 10,000 1.08P = 10,000 P = $9259.26 1.19 80,000 + 80,000(i) = 100,000 i = 25% 1.22 Simple: 1,000,000 = 500,000 + 500,000(i)(5) i = 20% per year simple Compound: 1,000,000 = 500,000(1 + i)5 (1 + i)5 = 2.0000 (1 + i) = (2.0000)0.2 i = 14.87%

1.25 Plan 1: Interest paid each year = 400,000(0.10) = $40,000 Chapter 1 1

Total paid = 40,000(3) + 400,000 = $520,000 Plan 2: Total due after3 years = 400,000(1 + 0.10)3 = $532,400 Difference paid = 532,400 – 520,000 = $12,400 1.28 (a) FV(i%,n,A,P) finds the future value, F (b) IRR(first_cell:last_cell) finds the compound interest rate, i (c) PMT(i%,n,P,F) finds the equal periodic payment, A (d) PV(i%,n,A,F) finds the present value, P. 1.31 For built-in Excel functions, a parameter that does not apply can be left blank when it is notan interior one. For example, if there is no F involved when using the PMT function to solve a particular problem, it can be left blank because it is an end function. When the function involved is an interior one (like P in the PMT function), a comma must be put in its position. 1.34 Highest to lowest rate of return is as follows: Credit card, bank loan to new business, corporate bond,government bond, interest on checking account 1.37 End of period convention means that the cash flows are assumed to have occurred at the end of the period in which they took place. 1.40 The cash flow diagram is:
P=?
i = 15%

0 1

2

3

4

5

$40,000

1.43 4 = 72/i i = 18% per year 1.46 2P = P + P(0.05)(n) n = 20 Answer is (d) 1.49 Answer is (c)

Chapter 1

2

SOLUTIONS TO SELECTEDPROBLEMS Student: You should work the problem completely before referring to the solution.

CHAPTER 2
Solutions included for problems 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, and 82 2.1 1. (F/P,8%25) = 6.8485; 2. (P/A,3%,8) = 7.0197; 3. (P/G,9%,20) = 61.7770; 4. (F/A,15%,18) = 75.8364; 5. (A/P,30%,15) = 0.30598 2.4 P =600,000(P/F,12%,4) = 600,000(0.6355) = $381,300 2.7 P = 75(P/F,18%,2) = 75(0.7182) = $53.865 million 2.10 P = 162,000(P/F,12%,6) = 162,000(0.5066) = $82,069 2.13 P = 1.25(0.10)(P/F,8%,2) + 0.5(0.10)(P/F,8%,5) = 0.125(0.8573) + 0.05(0.6806) = $141,193 2.16 A = 1.8(A/P,12%,6) = 1.8(0.24323) = $437,814 2.19 P = 75,000(P/A,15%,5) = 75,000(3.3522) = $251,415 2.22 P = 2000(P/A,8%,35) = 2000(11.6546) = $23,3092.25 (a) 1. Interpolate between n = 32 and n = 34: 1/2 = x/0.0014 x = 0.0007 (P/F,18%,33) = 0.0050 – 0.0007 = 0.0043

Chapter 2

1

2. Interpolate between n = 50 and n = 55: 4/5 = x/0.0654 x = 0.05232 (A/G,12%,54) = 8.1597 + 0.05232 = 8.2120 (b) 1. (P/F,18%,33) = 1/(1+0.18)33 = 0.0042 2. (A/G,12%,54) = {(1/0.12) – 54/[(1+0.12)54 –1} = 8.2143 2.28 (a) G = $5 million (b) CF6 = $6030 million (c)n = 12

2.31 (a) CF3 = 280,000 – 2(50,000) = $180,000 (b) A = 280,000 – 50,000(A/G,12%,5) = 280,000 – 50,000(1.7746) = $191,270 2.34 A = 14,000 + 1500(A/G,12%,4) = 14,000 + 1500(1.3589) = $16,038 2.37 50 = 6(P/A,12%,6) + G(P/G,12%,6) 50 = 6(4.1114) + G(8.9302) G = $2,836,622 2.40 For g = i, P = 60,000(0.1)[15/(1 + 0.04)] = $86,538 2.43 First find P and then convert to F: P = 2000{1 –...
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